Question

Small distances are commonly measured capacitively. Consider an air-filled parallel-plate capacitor with fixed plate area A=82mm² and a variable plate-separation distance x. Assume this capacitor is attached to a capacitance-measuring instrument which can measure capacitance C in the range 3.0pF to 3000.0pF with an accuracy of 0.6pF .

A)

If C is measured while x is varied, over what range (xmin <= x <= xmax) can the plate-separation distance (in um) be determined by this setup?

B) Define Δx to be the accuracy (magnitude) to which x can be determined, and determine a formula for  Δx.

Express your answer in terms of the variables x,  ΔC, A, and appropriate constants.

D)

Determine the percent accuracy to which xmin and xmax can be measured.

Answer 1

A) the capacitance of parrallel plate capacitor is given by

   C= $\epsilon$ ₀A/x

   x= $\epsilon$ ₀A/C

   Now putting the values in the given equation

xmax=(8.85 $\times$ 10^-12)(82 $\times$ (10^-3 )² / (3 $\times$ 1o^-12)              when C= 3pF

xmax=0.00024m

Now putting the C=3000pF , we get

xmin= o.ooo12m

B)   By accuracy we mean the relative error in measurement

     accuracy $\times$ measured quantity=error

      From above capacitance value

      o.6 $\times$ 3pF=1.8

      and Now the difference between actual value and measured value is

      Now     error= actual value - measurd value

                  actual value= error+measured vlue

                  actual Capacitance= 1.8 + 3= 3.8pF

         Now for the actual value of Capacitance the sepration x will be

               xact= ( 8.85 $\times$ 10^-12) (82 $\times$ (10^-3)² / (4 $\times$ 10^-12)

                xact=0.00018

error in x= x_act - x_mea

                =o.ooo18-0.00024=-0.000058m

error in x =0.000058m

Again using the formula of accurracy

accuracy=error / measure quantity

               =0.000058/0.00024

accuracy =0.244m

So by using this tecqnique you can find the accuracy for minium sepration value.

D)            For percentage accuracy you got to find the percentage of th\e value of the accuracy.

                   For the percent accuracy of x_min= 0.244 $\times$ 100/1oo

                                                                            =24.4%

                         By using same procedure you can find for the x_max